What is the energy of a photon emitted during the transition from n=3 to n=2 in

What is the energy of a photon emitted during the transition from n=3 to n=2 in a hydrogen atom?

What is the energy of a photon emitted during the transition from n=3 to n=2 in a hydrogen atom?

Correct Answer: 1.89 eV

Step 1: Understand that in a hydrogen atom, electrons can exist in different energy levels, which are represented by 'n' values (like n=1, n=2, n=3, etc.).
Step 2: Identify the initial and final energy levels for the transition. Here, the electron is moving from n=3 (higher energy level) to n=2 (lower energy level).
Step 3: Use the Rydberg formula to calculate the energy of the photon emitted during this transition. The formula is: E = 13.6 eV * (1/n_final^2 - 1/n_initial^2).
Step 4: Plug in the values into the formula: E = 13.6 eV * (1/2^2 - 1/3^2).
Step 5: Calculate 1/2^2 = 1/4 = 0.25 and 1/3^2 = 1/9 ≈ 0.1111.
Step 6: Subtract these values: 0.25 - 0.1111 = 0.1389.
Step 7: Multiply by 13.6 eV: E ≈ 13.6 eV * 0.1389 ≈ 1.89 eV.
Step 8: Conclude that the energy of the photon emitted during the transition from n=3 to n=2 is approximately 1.89 eV.

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